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Class XI · Physics

Laws of Motion: NEET MCQ practice

283 practice questions on Laws of Motion, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

283

Practice questions

175

Easy

105

Medium

3

Hard

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What Laws of Motion covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Force and inertia, Newton's laws of motion
  • Momentum, impulse, conservation of linear momentum
  • Equilibrium of concurrent forces
  • Static and kinetic friction, laws of friction, rolling friction
  • Dynamics of uniform circular motion (vehicle on level/banked road)

Sample questions on Laws of Motion

Try each one, then open the answer and explanation.

Question 1 · Practice question

An iron block of sides 50 cm x 8 cm x 15 cm has to be pushed along the lloor. The force required will be niinimum(when) the surface in contact with ground is

  1. A.8 cm x 15 cm surface
  2. B.50 cm x 15 cm surface
  3. C.8 cm x 50 cm surface
  4. D.force is same for all surfaces
Show answer

Answer: D. force is same for all surfaces

The force needed to push the block is the frictional force, which is given by μN, where N = mg is constant. According to the laws of friction, for solid surfaces, kinetic friction is independent of the area of contact. Therefore, no matter which face is in contact with the ground, the force required remains the same. Option (d) is correct; the tempting wrong options (a, b, c) might trick students who incorrectly think that pressure (force per unit area) affects frictional force.

Question 2 · Practice question

A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m s -2. The total mass of the crew and passengers is 500 kg. Choose the correct statements from the following. (Take g = 10 m s -2 ) (i) The force on the floor of the helicopter by the crew and passengers is 1.25 x 10 4 N vertically downwards. (ii) The action of the rotor of the helicopter on the surrounding air is 6.25 x 10 4 vertically downwards. (iii) Theforceon the helicopter due to the surrounding air is 6.25 x 10 4 N vertically upwards.

  1. A.(i) and (ii)
  2. B.(ii) and (iii)
  3. C.(i) and (iii)
  4. D.All the three
Show answer

Answer: D. All the three

Correct because: For (i), crew experience upward acceleration so normal force exceeds weight, giving 1.25×10⁴ N downward force on floor by Newton’s third law; for (ii) and (iii), upward force by air on helicopter equals total mass times (g+a) = 6.25×10⁴ N, making rotor’s action on air equal and opposite downward (ii), and force on helicopter upward (iii). The common trap is confusing action–reaction directions or misreading whose force is whose.

Question 3 · Practice question

A person in an elevator accelerating upwards with an acceleration of 2 m s -2, tosses a coin vertically upwards with a speed of 20 m s -1. After how much time will the coin fall back into his hand? (Take g = 10 m s -2 )

  1. A.5/3s
  2. B.3/10s
  3. C.10/3s
  4. D.3/5s
Show answer

Answer: C. 10/3s

In the accelerating elevator frame, the effective acceleration acting downwards is g(eff) = g + a = 10 + 2 = 12 m/s², because upward acceleration increases the apparent weight. Time of flight formula T = 2u / g(eff) gives T = 2×20 / 12 = 10/3 s. A common mistake is to simply use g = 10 m/s² without adjusting for the elevator's acceleration, which would give 4 s — not among the options, so many would be forced to rethink.

Question 4 · Practice question

A person of mass 50 kg stands on a weighing scale on a lift. If the lift is ascending upwards with a uniform acceleration o f 9 m s -2, what would be the reading of the weighing scale? (Take g = 10m s -2 )

  1. A.50 kg
  2. B.60 kg
  3. C.95 kg
  4. D.100 kg
Show answer

Answer: C. 95 kg

When the lift accelerates upward, the scale reads a larger value because the normal force must also provide the upward acceleration. Using N = m(g + a), with m = 50 kg, g = 10 m/s², and a = 9 m/s², we get N = 950 N, which corresponds to 95 kg on a scale calibrated for g = 10 m/s². The tempting wrong option 100 kg would be correct if the acceleration were 10 m/s² instead of 9 m/s², but careful substitution shows only 95 kg matches.

Question 5 · Practice question

Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is

  1. A.g/4
  2. B.g/5
  3. C.g/8
  4. D.g/6
Show answer

Answer: B. g/5

The acceleration of an Atwood machine system is given by a = (m2 − m1)g / (m1 + m2). Substituting m1 = 8 kg and m2 = 12 kg yields a = (12 − 8)g / (8 + 12) = 4g/20 = g/5. Wrong options come from mis‑remembering the formula (like swapping numerator and denominator) or simple arithmetic mistakes, but the direct recall of this standard result leads quickly to option B.

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